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(a) Express the second law of thermodynamics in words. (b) If the entropy of the system increases during a reversible process, what can you say about the entropy change of the surroundings? (c) In a certain spontaneous process the system undergoes an entropy change, \(\Delta S=42 \mathrm{~J} / \mathrm{K} .\) What can you conclude about ?

Short Answer

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(a) The Second Law of Thermodynamics states that the total entropy of an isolated system can never decrease over time; it can only increase or remain constant during a reversible process. (b) If the entropy of the system increases for a reversible process, then the entropy of the surroundings must have decreased by the same amount, i.e., ∆S_surroundings = -∆S_system. (c) For the given spontaneous process with ∆S = 42 J/K, the entropy change of the surroundings (∆S_surroundings) will be greater than -42 J/K, i.e., more positive than -42 J/K, to maintain a net increase in the total entropy of the universe.

Step by step solution

01

(a) Second Law of Thermodynamics in words

The Second Law of Thermodynamics states that the total entropy of an isolated system can never decrease over time; it can only increase or remain constant during a reversible process. It basically means that natural systems tend to move towards a state of disorder or randomness.
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(b) Entropy change in surroundings for a reversible process

During a reversible process, the system is maintained in thermodynamic equilibrium with its surroundings, and no energy is lost in the form of heat. If the entropy of the system increases for a reversible process, then the entropy of the surroundings must have decreased by the same amount. This is because the total entropy change (∆S) for a reversible process is equal to the sum of the entropy changes of the system (∆S_system) and the surrounding (∆S_surroundings): ∆S_total = ∆S_system + ∆S_surroundings Since ∆S_total = 0 for a reversible process, ∆S_surroundings = -∆S_system.
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(c) Concluding based on given spontaneous process

In this spontaneous process, the entropy change of the system (∆S) is given as 42 J/K. Since it is a spontaneous process, it must result in a net increase in the total entropy of the universe. Therefore, we can conclude that the entropy change of the surroundings (∆S_surroundings) will be greater than -42 J/K, which in other words means that the entropy change of the surroundings will be more positive than -42 J/K. This is because, for a spontaneous process ∆S_total > 0, the decrease in entropy of the surroundings must be less than the increase in entropy of the system.

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Key Concepts

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

Entropy
Entropy is a fundamental concept in thermodynamics, often dubbed the measure of disorder or randomness in a system. More formally, entropy quantifies the number of microstates—essentially, the possible configurations that a system can have at a given energy level. The Second Law of Thermodynamics, interpreted in terms of entropy, suggests that in an isolated system, entropy can either stay constant or increase over time; it never decreases.
As entropy increases, a system becomes more disordered, and less energy is available to do work. This has broad implications, particularly in understanding energy conversion processes and the direction of chemical reactions. For example, when a piece of ice melts into water, the entropy increases since the liquid state has more possible configurations than the solid phase. Likewise, when gas expands in a container, the entropy of the gas increases due to an increase in the number of potential microstates.

Importance of Entropy in Various Fields

  • Thermodynamics: Determines the feasibility and directionality of reactions.
  • Statistical Mechanics: Connects the microscopic interactions with macroscopic observations.
  • Information Theory: Analogous to the degree of surprise or uncertainty in data.
  • Cosmology: Helps understand the arrow of time and the evolution of the universe.
It's imperative to comprehend that while the entropy of a system may decrease, the universe's total entropy never decreases; instead, it stays constant or increases, as described by the Second Law. This overarching principle governs everything from steam engines to the stars in the night sky.
Reversible Process
A reversible process in thermodynamics is an idealized concept, where a system changes in such a way that the system and environment can be restored to their original states without leaving any net change. In reality, no process is perfectly reversible; however, this concept is valuable for analyzing ideal behaviors and understanding the limits of efficiency.
During a reversible process, conditions must change infinitesimally slowly, allowing the system to remain in a state of equilibrium. An example of a reversible process is the very slow compression or expansion of a gas. If we imagine a piston compressing a gas very slowly, each infinitesimal step could theoretically be reversed by an equally infinitesimal release of the piston.

Characteristics of Reversible Processes

  • Equilibrium: The system is in a constant state of equilibrium with its surroundings.
  • No Energy Loss: There is no net loss or gain in energy as heat or work during a reversible process.
  • Maximal Efficiency: Reversible processes are the most efficient processes possible in thermodynamics.
  • Bidirectionality: A reversible process can move both 'forward' and 'backward' along its path.
Entropy changes in a reversible process must be counterbalanced by the surrounding system, maintaining the total entropy. Therefore, if a system undergoes a reversible increase in entropy, the surrounding environment must simultaneously experience an equivalent decrease in entropy, keeping the universe’s total entropy conserved.
Spontaneous Process
A spontaneous process is nature's tendency to move towards a state of equilibrium without the need for external energy. This means a process will 'naturally' happen if left alone. Spontaneity in thermodynamics is largely determined by changes in entropy. A spontaneous process generally leads to an increase in the entropy of the universe, which encompasses both the system of interest and its surroundings.
Unlike reversible processes, spontaneous processes are irreversible; once they occur, the system cannot return to its initial state without external influence. Common examples include ice melting at room temperature, salt dissolving in water, and a hot object cooling down to room temperature.

Cues for Predicting Spontaneity

  • Energy Distribution: Energy tends to disperse, leading to a more uniform spread throughout the system.
  • Phase Changes: Transitions from ordered (solid) to disordered (liquid/gas) phases are spontaneous.
  • Chemical Reactions: Reactions leading to an increase in gaseous products or dissolving solutes.
The given exercise illustrates a common misunderstanding: the belief that a spontaneous process might mean an instantaneous or rapid occurrence. In thermodynamics, 'spontaneous' simply implies that the process is energetically favorable and will occur without external assistance over some interval of time, which can be long. As the exercise solution outlines, if we know the entropy of the system increases by 42 J/K in a spontaneous process, the surrounding’s entropy decrease is not as much, meaning the total entropy of the universe increases, aligning well with the Second Law of Thermodynamics.

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Most popular questions from this chapter

In each of the following pairs, which compound would you expect to have the higher standard molar entropy: (a) \(\mathrm{C}_{2} \mathrm{H}_{2}(g)\) or \(\mathrm{C}_{2} \mathrm{H}_{6}(g)\) (b) \(\mathrm{CO}_{2}(g)\) or \(\mathrm{CO}(g) ?\) Explain.

Use data in Appendix \(\mathrm{C}\) to calculate \(\Delta H^{\circ}, \Delta S^{\circ},\) and \(\Delta G^{\circ}\) at \(25^{\circ} \mathrm{C}\) for each of the following reactions. In each case show that \(\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ}\) (a) \(2 \mathrm{Cr}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CrO}_{3}(s)\) (b) \(\mathrm{BaCO}_{3}(s) \longrightarrow \mathrm{BaO}(s)+\mathrm{CO}_{2}(g)\) (c) \(2 \mathrm{P}(s)+10 \mathrm{HF}(g) \longrightarrow 2 \mathrm{PF}_{5}(g)+5 \mathrm{H}_{2}(g)\) (d) \(\mathrm{K}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{KO}_{2}(s)\)

A certain reaction has \(\Delta H^{\circ}=+23.7 \mathrm{~kJ}\) and \(\Delta S^{\circ}=\) \(+52.4 \mathrm{~J} / \mathrm{K} .\) (a) Is the reaction exothermic or endothermic? (b) Does the reaction lead to an increase or decrease in the randomness or disorder of the system? (c) Calculate \(\Delta G^{\circ}\) for the reaction at \(298 \mathrm{~K} .(\mathbf{d})\) Is the reaction spontaneous at \(298 \mathrm{~K}\) under standard conditions?

The conversion of natural gas, which is mostly methane, into products that contain two or more carbon atoms, such as ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\), is a very important industrial chemical process. In principle, methane can be converted into ethane and hydrogen: $$ 2 \mathrm{CH}_{4}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)+\mathrm{H}_{2}(g) $$ In practice, this reaction is carried out in the presence of oxygen: $$ 2 \mathrm{CH}_{4}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$ (a) Using the data in Appendix \(C,\) calculate \(K\) for these reactions at \(25^{\circ} \mathrm{C}\) and \(500{ }^{\circ} \mathrm{C}\). (b) Is the difference in \(\Delta G^{\circ}\) for the two reactions due primarily to the enthalpy term \((\Delta H)\) or the entropy term \((-T \Delta S) ?(\mathbf{c})\) Explain how the preceding reactions are an example of driving a nonspontaneous reaction, as discussed in the "Chemistry and Life" box in Section 19.7 . (d) The reaction of \(\mathrm{CH}_{4}\) and \(\mathrm{O}_{2}\) to form \(\mathrm{C}_{2} \mathrm{H}_{6}\) and \(\mathrm{H}_{2} \mathrm{O}\) must be carried out carefully to avoid a competing reaction. What is the most likely competing reaction?

Consider what happens when a sample of the explosive TNT (Section 8.8: "Chemistry Put to Work: Explosives and Alfred Nobel") is detonated under atmospheric pressure. (a) Is the detonation a spontaneous process? (b) What is the sign of \(q\) for this process? (c) Can you determine whether \(w\) is positive, negative, or zero for the process? Explain. (d) Can you determine the sign of \(\Delta E\) for the process? Explain.

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