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A reaction occurs spontaneously if (a) \(\mathrm{T} \Delta \mathrm{S}<\Delta \mathrm{H}\) and both \(\Delta \mathrm{H}, \Delta \mathrm{S}\) are \(+\mathrm{ve}\) (b) \(\mathrm{T} \Delta \mathrm{S}>\Delta \mathrm{H}\) and \(\Delta \mathrm{H}=+\mathrm{ve}, \Delta \mathrm{S}=-\mathrm{ve}\) (c) \(\mathrm{T} \Delta \mathrm{S}>\Delta \mathrm{H}\) and both \(\Delta \mathrm{H}, \Delta \mathrm{S}\) are \(+\mathrm{ve}\) (d) \(\mathrm{T} \Delta \mathrm{S}=\Delta \mathrm{H}\) and both \(\Delta \mathrm{H}, \Delta \mathrm{S}\) are \(+\mathrm{ve}\)

Short Answer

Expert verified
Option (c) is correct: \(T \Delta S > \Delta H\) and both \(\Delta H, \Delta S\) are positive.

Step by step solution

01

Understanding Spontaneity Criteria

A reaction is spontaneous if the change in Gibbs free energy (9G) is negative. The formula for Gibbs free energy is \[\Delta G = \Delta H - T\Delta S\]where \(\Delta H\) is the change in enthalpy, \(T\) is the temperature, and \(\Delta S\) is the change in entropy. For spontaneity, we require \[\Delta G < 0.\]Therefore, \[T\Delta S > \Delta H.\]
02

Analyzing Option (a)

For option (a), the condition given is \(T \Delta S < \Delta H\) and both \(\Delta H\) and \(\Delta S\) are positive. This implies that\[\Delta G = \Delta H - T\Delta S > 0.\]This would not lead to a spontaneous reaction; hence option (a) is incorrect.
03

Analyzing Option (b)

In option (b), \(T \Delta S > \Delta H\) with \(\Delta H\) positive and \(\Delta S\) negative. Here,\[T\Delta S\text{ is negative, so }\Delta G = \Delta H - T\Delta S \geq \Delta H.\]With positive \(\Delta H\), \(\Delta G\) is positive, indicating non-spontaneity, so option (b) is also incorrect.
04

Analyzing Option (c)

For option (c), the condition is \(T \Delta S > \Delta H\) with both \(\Delta H\) and \(\Delta S\) positive. The calculation is:\[\Delta G = \Delta H - T\Delta S < 0.\]This leads to spontaneity; hence, option (c) is a valid scenario for a spontaneous reaction.
05

Analyzing Option (d)

Option (d) proposes \(T \Delta S = \Delta H\) with both \(\Delta H\) and \(\Delta S\) positive. This yields:\[\Delta G = \Delta H - T\Delta S = 0.\]This scenario is characteristic of equilibrium, not spontaneity. Thus, option (d) is not correct.

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Key Concepts

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

Spontaneous Reactions
Spontaneous reactions are processes that occur without the need for any external energy input. They are naturally inclined to happen and can be predicted using the concept of Gibbs Free Energy. For a chemical reaction to be spontaneous, the change in Gibbs Free Energy (\( \Delta G \)) must be negative. The equation that defines this relationship is \[ \Delta G = \Delta H - T\Delta S \] where:
  • \(\Delta H\) is the enthalpy change.
  • \(T\) is the temperature in Kelvin.
  • \(\Delta S\) is the entropy change.
This equation implies that even if a reaction has positive changes in enthalpy and entropy, it can still be spontaneous, provided that the term \(T\Delta S\) is greater than \(\Delta H\). This helps in determining not only if a reaction is spontaneous under certain conditions but also understanding how temperature can influence reaction spontaneity.
Entropy Change (ΔS)
Entropy is a measure of disorder or randomness in a system. The change in entropy, denoted as \(\Delta S\), reflects the increase or decrease in disorder during a reaction. Generally, systems tend to move towards higher entropy.

In a reaction, if \(\Delta S\) is positive, it signifies an increase in disorder, which often favors spontaneity, especially when temperature is a factor.
  • A positive \(\Delta S\) suggests products are more disordered than reactants, aligning with the natural tendency of systems toward chaos.
  • Conversely, a negative \(\Delta S\) indicates a decrease in disorder, making spontaneity less likely unless compensated by other factors like enthalpy.
The interplay between entropy and temperature is critical. Since temperature amplifies the effect of \(\Delta S\) in the Gibbs equation, even slight positive entropy changes can drive reactions to be spontaneous at higher temperatures.
Enthalpy Change (ΔH)
Enthalpy change, represented as \(\Delta H\), tells us about the heat absorbed or released in a reaction at constant pressure.
  • When \(\Delta H\) is negative, the reaction is exothermic, releasing energy to the surroundings, which often supports spontaneity.
  • If \(\Delta H\) is positive, the reaction is endothermic, absorbing energy, which can reduce the likelihood of spontaneity unless entropy changes or temperature play a significant role.
However, it is important to remember that enthalpy alone does not determine spontaneity. A crucial combination of enthalpy and entropy changes, along with temperature, jointly influence the reaction's spontaneity as per the Gibbs Free Energy equation. Therefore, in scenarios where both \(\Delta H\) and \(\Delta S\) are positive, the reaction might become spontaneous at a sufficient temperature.

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Most popular questions from this chapter

The enthalpy of vaporization of a liquid is \(30 \mathrm{~kJ} \mathrm{~mol}^{-1}\) and entropy of vaporization is \(5 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K} .\) The boiling point of the liquid at 1 atm is (a) \(250 \mathrm{~K}\) (b) \(400 \mathrm{~K}\) (c) \(450 \mathrm{~K}\) (d) \(600 \mathrm{~K}\)

Plants and living beings are examples of (a) isolated system (b) adiabatic system (c) open system (d) closed system

Identify the correct statement regarding a spontaneous process. (a) Endothermic processes are never spontaneous (b) Exothermic process are always spontaneous (c) Lowering of energy in the reaction process is the only criterion for spontaneity (d) For a spontaneous process in an isolated system, the change in entropy is positive.

In the reaction: \(\mathrm{CO}(\mathrm{g})+1 / 2 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{~g})\) the change in \(\Delta \mathrm{S}^{\circ}\) is (given \(\mathrm{S}^{\circ}\) for \(\mathrm{CO}, \mathrm{O}_{2}\) and \(\mathrm{CO}_{2}\) are \(197.6,205.3\) and \(213.6 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\) respectively) (a) \(-78.6 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\) (b) \(-50 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\) (c) \(-86.5 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\) (d) \(-30 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\)

Which one of the following statements is false? (a) temperature is a state function (b) work is a state function (c) change in the state depends upon initial and final state (d) work appears at the boundary of the system

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