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(a) In a chemical reaction two gases combine to form a solid. What do you expect for the sign of \(\Delta S ?\) (b) How does the entropy of the system change in the processes described in Exercise \(19.12 ?\)

Short Answer

Expert verified
\( \) a) When two gases combine to form a solid, the change in entropy, \(\Delta S\), will be negative (\(\Delta S < 0\)), as the final state (solid) has lower entropy than the initial state (two gases). b) In the process of charging a capacitor, there is a decrease in entropy for the surrounding air due to energy transfer, thus \(\Delta S < 0\).

Step by step solution

01

Analyze the change in particles and phase for part (a)

For this chemical reaction, two gases are combining to form a solid. Since gases have more random motions and disorder compared to solids, the products will have less randomness and order than the reactants. We can analyze this change by comparing the number of particles and their phases before and after the chemical reaction.
02

Determine the sign of ΔS for part (a)

Considering the decrease in randomness and order as gases combine to form a solid, the change in entropy of the system, ΔS, will be negative. This is because the final state (solid) has lower entropy than the initial state (two gases). Thus, for part (a) we can conclude: ΔS < 0 Now let's analyze part (b).
03

Recall the process from Exercise 19.12

Exercise 19.12 states: "Work is done in charging a capacitor. How does this energy transfer affect the entropy of the surrounding air?" To answer part (b), we need to analyze how charging a capacitor and the energy transfer will affect the entropy of the surrounding air.
04

Analyze the energy transfer while charging a capacitor

When work is done to charge a capacitor, the energy is stored as electrical potential energy in the capacitor. This process requires energy transfer from the surroundings to the capacitor. Since the surroundings are losing energy, this will cause a decrease in the entropy of the surrounding air.
05

Determine the entropy change for part (b)

As we have concluded that the energy transfer will lead to a decrease in entropy for the surrounding air, we can express the change in entropy of the system for part (b) as: ΔS < 0

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Key Concepts

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

Entropy Change in Reactions
Entropy, symbolized by the letter S, is a measure of randomness or disorder in a system. When a chemical reaction occurs, entropy can either increase or decrease depending on the state of the reactants and products. In the case of gases combining to form a solid, as mentioned in the exercise, the disorder of the system decreases. Gases have more freedom of movement and therefore higher entropy than solids, which have a fixed structure and less randomness. During the reaction, as the particles come together to form a more ordered solid, the overall entropy decreases. Consequently, the symbol \( \Delta S \) represents the change in entropy, and in this scenario, we can affirm that \( \Delta S < 0 \)— which indicates a negative change in entropy. This concept is critical to understanding the direction of chemical reactions and predicting the spontaneous nature of processes in thermodynamics.

When attempting to grasp the intricacies of entropy change in reactions, it is essential to evaluate the states of matter of both the reactants and products, alongside considering energy changes during the reaction. For instance, a reaction that produces more gaseous molecules from fewer will typically result in an increase in entropy, since gases have higher entropy than liquids or solids. Similarly, a solution becoming more disordered or mixed would lead to higher entropy, often seen in dissolving or diffusion processes.
Phase Changes and Entropy
Phase changes are physical transitions between different states of matter: gas, liquid, and solid. These changes are associated with corresponding alterations in entropy. During a phase change, energy is either absorbed from the surroundings — endothermic process — or released into the surroundings — exothermic process. An endothermic transition, such as melting or vaporization, leads to an increase in entropy because the particles have more freedom in the liquid or gas phase compared to the solid phase. Conversely, exothermic transitions, like freezing or condensation, result in a decrease in entropy due to the reduced movement and increased order of particles.

Understanding phase changes is crucial for comprehending entropy variations within a system. A useful rule of thumb is that the transition from solid to liquid to gas typically leads to an increase in entropy. This information helps predict the final state of a system and the feasibility of phase transitions. For example, the conversion of water from ice to liquid water to steam is accompanied by a clear increase in entropy due to the particles becoming progressively less ordered and more dispersed.
Energy Transfer and Entropy
In thermodynamics, energy transfer is a key determinant in changes of entropy. When energy is transferred to or from a system, it affects the molecular motion and randomness of that system. For example, as in exercise 19.12, when energy is transferred from the surrounding air to charge a capacitor, the air cools down slightly. Less energy in the air means less molecular motion, leading to lower entropy in the surroundings. Conversely, if the energy were transferred from the capacitor to the air, the increased movement of air molecules would result in an increase in entropy.

The concept of energy conservation also plays a pivotal role here; while the total energy of an isolated system remains constant, the way it is distributed can affect the entropy of the system and its surroundings. Energy transfer through heat is the most common cause of changes in entropy. For example, when a substance absorbs heat, the added energy increases molecular motion and consequently increases the disorder, thereby increasing entropy. Understanding these principles is essential for grasping the second law of thermodynamics, which states that the total entropy of a closed system can never decrease over time — it can either increase or remain unchanged, a concept known as entropy production or generation.

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Most popular questions from this chapter

Methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) can be made by the controlled oxidation of methane: $$ \mathrm{CH}_{4}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(g) $$ (a) Use data in Appendix C to calculate \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for this reaction. (b) How is \(\Delta G^{\circ}\) for the reaction expected to vary with increasing temperature? (c) Calculate \(\Delta G^{\circ}\) at \(298 \mathrm{~K}\). Under standard conditions, is the reaction spontaneous at this temperature? (d) Is there a temperature at which the reaction would be at equilibrium under standard conditions and that is low enough so that the compounds involved are likely to be stable?

(a) Express the second law of thermodynamics as a mathematical equation. (b) In a particular spontaneous process the entropy of the system decreases. What can you conclude about the sign and magnitude of \(\Delta S_{\text {surr }} ?\) (c) During a certain reversible process, the surroundings undergo an entropy change, \(\Delta S_{\text {surr }}=-78 \mathrm{~J} / \mathrm{K}\). What is the entropy change of the system for this process?

For the majority of the compounds listed in Appendix \(\mathrm{C},\) the value of \(\Delta G_{f}^{\circ}\) is more positive (or less negative) than the value of \(\Delta H_{f}^{\circ} .\) (a) Explain this observation, using \(\mathrm{NH}_{3}(g), \mathrm{CCl}_{4}(l)\), and \(\mathrm{KNO}_{3}(s)\) as examples. (b) An exception to this observation is \(\mathrm{CO}(g)\). Explain the trend in the \(\Delta H_{f}^{\circ}\) and \(\Delta G_{f}^{\circ}\) values for this molecule.

The crystalline hydrate \(\mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2} \cdot 4 \mathrm{H}_{2} \mathrm{O}(s)\) loses water when placed in a large, closed, dry vessel: $$ \mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2} \cdot 4 \mathrm{H}_{2} \mathrm{O}(s) \longrightarrow \mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2}(s)+4 \mathrm{H}_{2} \mathrm{O}(g) $$ This process is spontaneous and \(\Delta H\) is positive. Is this process an exception to Bertholet's generalization that all spontaneous changes are exothermic? Explain.

The standard entropies at \(298 \mathrm{~K}\) for certain of the group \(4 \mathrm{~A}\) elements are as follows: \(\mathrm{C}(s,\) diamond \()=2.43 \mathrm{~J} / \mathrm{mol}-\mathrm{K},\) \(\mathrm{Si}(s)=18.81 \mathrm{~J} / \mathrm{mol}-\mathrm{K}, \mathrm{Ge}(s)=31.09 \mathrm{~J} / \mathrm{mol}-\mathrm{K},\) and \(\operatorname{Sn}(s)=\) \(51.818 \mathrm{~J} / \mathrm{mol}-\mathrm{K} .\) All but \(\mathrm{Sn}\) have the diamond structure. How do you account for the trend in the \(S^{\circ}\) values?

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