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Chapter 4: Problem 92

Which of the following conditions will always lead to a non-spontaneous, change? (a) \(\Delta H\) and \(\Delta S\) both \(+\) ve (b) \(\Delta H\) is -ve and \(\Delta S\) is \(+\) ve (c) \(\Delta H\) and \(\Delta S\) both \(-v e\) (d) \(\Delta H\) is +ve and \(\Delta S\) is -ve

Short Answer

Expert verified
Condition (d) H is positive and S is negative will always lead to a positive G, indicating a non-spontaneous process.

Step by step solution

01

- Understand Spontaneity and the Second Law of Thermodynamics

The spontaneity of a reaction is determined by the Gibbs free energy change (G). According to the second law of thermodynamics, for a process to be spontaneous, G must be negative. The Gibbs free energy is defined as G = H - TS, where H is the change in enthalpy, T is the temperature in Kelvin, and S is the change in entropy.
02

- Determine the Effect of Conditions on G

Examine the sign of H and S under each given condition to determine its effect on G. A positive H will increase G, while a negative H will decrease G. A positive S will tend to decrease G, while a negative S will increase it since S is multiplied by the negative temperature (T).
03

- Apply Conditions to the G Equation

By substituting the signs of H and S into the G equation, determine under what conditions G is positive, which would indicate a non-spontaneous process. The condition where H is positive and S is negative will always lead to G being positive because the positive value of H will not be offset by the negative value of TS.

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Key Concepts

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

Thermodynamics
Thermodynamics is a branch of physics that deals with heat, energy, and the relationships between them.
It's fundamentally concerned with the flow and transformation of energy within physical systems, and it's vital in predicting how systems respond to changes in their environment.
Understanding the basic laws of thermodynamics can help explain why certain processes occur spontaneously.
Often, we're concerned with systems exchanging heat and work with their surroundings, and how this affects the properties of the system, leading to concepts such as enthalpy and entropy.
Enthalpy Change (\(\triangle H\))
Enthalpy change, often expressed as \(\triangle H\), is a measure of the total energy of a thermodynamic system, including both internal energy and the energy required to make room for it by displacing its environment.
In the context of chemical reactions, \(\triangle H\) indicates the heat absorbed or released under constant pressure.
If \(\triangle H\) is negative, the reaction is exothermic, releasing heat to the surroundings, and if positive, the reaction is endothermic and absorbs heat from the surroundings.
For a process to contribute to spontaneity, a release of heat (egative \(\triangle H\)) is often favorable.
Entropy Change (\(\triangle S\))
Entropy change, represented as \(\triangle S\), measures the disorder or randomness in a system.
It's a fundamental concept as it helps explain the direction in which chemical processes and physical processes take place.
An increase in entropy (positive \(\triangle S\)) favors spontaneity because systems naturally progress towards a state of higher randomness or disorder. Conversely, processes that result in a decrease in entropy (negative \(\triangle S\)) are typically non-spontaneous unless accompanied by other factors that can drive the process to occur.
Chemical Spontaneity
Chemical spontaneity refers to the inherent tendency of a chemical process to occur without external intervention.
The driving force behind this spontaneity is the decrease in Gibbs free energy, which incorporates both enthalpy and entropy changes in the context of temperature.
A spontaneous reaction will have a negative Gibbs free energy change, symbolizing that it can perform work on its surroundings, whereas a positive Gibbs free energy signifies a non-spontaneous reaction, which requires work to proceed.
Second Law of Thermodynamics
The second law of thermodynamics is a fundamental principle that states that the total entropy of an isolated system can never decrease over time.
It can only remain constant or increase, which implies that processes occur in a direction that increases the overall entropy of the universe.
This law is crucial to determining chemical spontaneity since it underpins the preference for processes that generate higher disorder.
Gibbs free energy is a direct application of this law, providing a quantitative measure that predicts the spontaneity of processes in systems not isolated but at constant temperature and pressure.

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Most popular questions from this chapter

Determine the enthalpy of formation of \(\mathrm{B}_{2} \mathrm{H}_{6}(g)\) in \(\mathrm{kJ} / \mathrm{mol}\) of the following reaction. $$ \mathrm{B}_{2} \mathrm{H}_{6}(g)+3 \mathrm{O}_{2}(g) \longrightarrow \mathrm{B}_{2} \mathrm{O}_{3}(s)+3 \mathrm{H}_{2} \mathrm{O}(g) $$ Given : \(\Delta_{r} H^{\circ}=-1941 \cdot \mathrm{kJ} / \mathrm{mol} ; \quad \Delta H_{f}^{\circ}\left(\mathrm{B}_{2} \mathrm{O}_{3}, s\right)=-1273 \mathrm{~kJ} / \mathrm{mol}\) \(\mathrm{W}_{f}^{\circ}\left(\mathrm{H}_{2} \mathrm{O}, g\right)=-241.8 \mathrm{~kJ} / \mathrm{mol}\) (a) \(-75.6\) (b) \(+7 \overline{5} .6\) (c) \(-57.4\) (d) \(-28.4\)

Calculate standard entropy change in the reaction $$ \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{H}_{2} \mathrm{O}(l) $$ Given : \(S_{m}^{\circ}\left(\mathrm{Fe}_{2} \mathrm{O}_{3}, \mathrm{~S}\right)=87.4, S_{m}^{\circ}(\mathrm{Fe}, S)=27.3\) \(S_{m}^{\circ}\left(\mathrm{H}_{2}, g\right)=130.7, S_{m}^{\circ}\left(\mathrm{H}_{2} \mathrm{O}, l\right)=69.9 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\)

Calculate the final temperature of a monoatomic ideal gas that is compressed reversible and adiabatically from \(16 \mathrm{~L}\) to \(2 \mathrm{~L}\) at \(300 \mathrm{~K}\) : (a) \(600 \mathrm{~K}\) (b) \(1044.6 \mathrm{~K}\) (c) \(1200 \mathrm{~K}\) (d) \(2400 \mathrm{~K}\)

One mole of an ideal gas at \(25^{\circ} \mathrm{C}\) expands in volume from \(1.0 \mathrm{~L}\) to \(4.0 \mathrm{~L}\) at constant . temperature. What work (in \(J\) ) is done if the gas expands against vacuum \(\left(P_{\text {cxtemal }}=0\right)\) ? (a) \(-4.0 \times 10^{2}\) (b) \(-3.0 \times 10^{2}\) (c) \(-1.0 \times 10^{2}\) (d) Zero

Calculate the entropy change \((\mathrm{J} / \mathrm{mol} \mathrm{K})\) of the given reaction. The molar entropies [J/K-mol] are given in brackets after each substance. \(2 \mathrm{PbS}(s)[91.2]+3 \mathrm{O}_{2}(g)[205.1] \longrightarrow 2 \mathrm{PbO}(s)[66.5]+2 \mathrm{SO}_{2}(g)[248.2]\) (a) \(-113.5\) (b) \(-168.3\) (c) \(+72.5\) (d) \(-149.2\)
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