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

Short Answer

Expert verified
(a) In a chemical reaction where two gases combine to form a solid, the sign of ΔS is expected to be negative, as the disorder decreases. (b) For both processes described in Exercise 19.12, the entropy of the system increases, and the sign of ΔS is expected to be positive.

Step by step solution

01

(a) Identify the initial and final states of the system

In this scenario, the initial state of the system consists of two gases, while the final state is a solid. Consider that gases have more disorder compared to solids.
02

(a) Determine the change in entropy for the reaction

As the system goes from two gases to a solid, there is a decrease in the disorder as the molecules become more organized in the solid state. Therefore, we expect the entropy change, ΔS, to be negative.
03

(a) Final conclusion for part (a)

In a chemical reaction where two gases combine to form a solid, the sign of ΔS is expected to be negative, as the disorder decreases.
04

(b) Review Exercise 19.12

In Exercise 19.12, the properties and processes of a system are described. For this part, we need to analyze how the entropy of the system changes in the processes mentioned in Exercise 19.12. Exercise 19.12 description: (i) A perfect gas expands in volume (reversibly) without any change in temperature. (ii) A perfect gas expands within a container due to a temperature increase.
05

(b) Analyze process (i)

In this process, a perfect gas expands in volume (reversibly) without any change in temperature. As the gas expands, the molecules are more spread out and occupy a larger volume, which leads to an increase in disorder. Hence, the entropy change, ΔS, should be positive.
06

(b) Analyze process (ii)

In this process, a perfect gas expands within a container due to a temperature increase. The expansion of the gas and the increase in temperature both contribute to an increase in disorder. Thus, the entropy change, ΔS, should be positive in this case as well.
07

(b) Final conclusion for part (b)

For both processes described in Exercise 19.12, the entropy of the system increases, and the sign of ΔS is expected to be positive.

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Key Concepts

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

Chemical Reaction
In a chemical reaction, substances interact to form new products, often accompanied by energy changes. These reactions can occur in a variety of ways, including the transformation of different states of matter. When two gases react to form a solid, there's a notable shift in the level of disorder in the system. Gases are highly disordered due to the free movement of their molecules. However, when they combine to form a solid, the molecules become tightly packed with a fixed arrangement. This transition from gas to solid results in a decrease in entropy, or disorder. Thus, for such a reaction, we can expect the change in entropy, denoted by ΔS, to be negative. In summary, the transformation involves a higher degree of organization, indicative of a reduction in entropy. Understanding this concept helps in predicting the behavior of substances in chemical reactions.
Gases and Solids
Gases and solids represent two distinct states of matter, characterized by differing molecular arrangements and levels of entropy.
  • Gases: Molecules in a gas are spaced far apart, moving freely and rapidly in all directions. This results in high entropy due to their random distribution.
  • Solids: In contrast, the molecules in solids are closely packed in a fixed, orderly structure. This ordered arrangement corresponds to low entropy.
When two gases react to form a solid, as seen in the chemical reaction example, there is a significant reduction in molecular freedom and, thus, a decrease in entropy. This illustrates the conservation principle in thermodynamics where systems tend to progress towards greater stability and lower energy levels.
Disorder and Order
Disorder and order describe the arrangement and energy state of molecules in a system. These concepts are foundational to understanding entropy, a key term in thermodynamics. Disorder, or high entropy, occurs when molecules are distributed randomly with significant freedom of movement, as in gases. Order, on the other hand, signifies low entropy, where molecules have limited movement and structured arrangement, typical of solids.Whenever a system transitions from a state of high disorder to higher order (or vice versa), the entropy changes. For example, when gases form a solid, the system moves towards more order, resulting in a decreased entropy (negative ΔS). Conversely, processes involving expansion in gases typically lead to increased disorder and positive entropy change, as seen in both scenarios described in Exercise 19.12, where gas molecules spread out more and become less ordered.
Thermodynamics
Thermodynamics is the study of energy, heat, and their transformations in physical and chemical processes. Entropy, a central concept in thermodynamics, measures the degree of molecular disorder within a system.
  • Entropy and Chemical Reactions: During a chemical reaction, changes in entropy indicate whether the disorder increases or decreases. A negative ΔS suggests that the system becomes more ordered, while a positive ΔS implies an increase in disorder.
  • Entropy in Gaseous State Changes: When gases expand or rise in temperature, as highlighted in Exercise 19.12, their entropy typically increases, reflecting a more randomized, spread-out molecular arrangement.
Understanding these principles allows us to predict the behavior of systems in various thermodynamic conditions and their drive towards equilibrium, where the entropy is maximized.

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Most popular questions from this chapter

Using the data in Appendix C and given the pressures listed, calculate Kp and ΔG for each of the following reactions:  (a) N2(g)+3H2(g)2NH3(g)PN2=2.6atm,PH2=5.9atm,RNH3=1.2atm (b) 2N2H4(g)+2NO2(g)3N2(g)+4H2O(g)PN2H4=PNO2=5.0×102atmPN2=0.5atm,PH2O=0.3atm (c) N2H4(g)N2(g)+2H2(g)PN2H4=0.5atm,PN2=1.5atm,PH2=2.5atm

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(a) Is the standard free-energy change, ΔG, always larger than ΔG?(b) For any process that occurs at constant temperature and pressure, what is the significance of ΔG=0 ? (c) For a certain process, ΔG is large and negative. Does this mean that the process, necessarily has a low activation barrier?

Reactions in which a substance decomposes by losing CO are called decarbonylation reactions. The decarbonylation of acetic acid proceeds according to: CH3COOH(l)CH3OH(g)+CO(g) By using data from Appendix C , calculate the minimum temperature at which this process will be spontaneous under standard conditions. Assume that ΔH and ΔS do not vary with temperature.

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