/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 43 Predict the sign of the entropy ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Predict the sign of the entropy change of the system for each of the following reactions: (a) \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)\) (b) \(\mathrm{CaCO}_{3}(s) \longrightarrow \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)\) (c) \(3 \mathrm{C}_{2} \mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{6}(g)\) (d) \(\mathrm{Al}_{2} \mathrm{O}_{3}(s)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{Al}(s)+3 \mathrm{H}_{2} \mathrm{O}(g)\)

Short Answer

Expert verified
(a) Negative entropy change. (b) Positive entropy change. (c) Negative entropy change. (d) Positive entropy change.

Step by step solution

01

a) N2(g) + 3 H2(g) → 2 NH3(g)

In this reaction, we have 4 moles of gaseous reactants transforming into 2 moles of gaseous products. The number of gas molecules is decreasing, which means that the entropy is decreasing. Therefore, the sign of the entropy change for this reaction is negative.
02

b) CaCO3(s) → CaO(s) + CO2(g)

In this reaction, we have one mole of solid reactant transforming into one mole of solid product and one mole of gaseous product. Since the reaction goes from a solid phase to a mixture of solid and gas phases, the entropy is increasing. Therefore, the sign of the entropy change for this reaction is positive.
03

c) 3 C2H2(g) → C6H6(g)

In this reaction, we have 3 moles of gaseous reactants transforming into 1 mole of gaseous product. The number of gas molecules is decreasing, which means that the entropy is decreasing. Therefore, the sign of the entropy change for this reaction is negative.
04

d) Al2O3(s) + 3 H2(g) → 2 Al(s) + 3 H2O(g)

In this reaction, three moles of gaseous reactants and one mole of solid reactant, with a total of 4 moles of reactants, transform into three moles of gaseous products and two moles of solid products, making a total of 5 moles of products. Since the number of gas molecules involved remains the same in this reaction, we will have to look at the overall increase in the number of particles. Here, the total number of particles increases from 4 moles to 5 moles. Therefore, the sign of the entropy change for this reaction is positive.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Chemical Thermodynamics
Chemical thermodynamics is the study of the interrelation of heat and work with chemical reactions or with physical changes of state within the confines of the laws of thermodynamics. It involves assessing the energy changes that occur during a process and how energy is transferred and converted.

One of the fundamental concepts in chemical thermodynamics is the first law, which states that energy cannot be created or destroyed, only transformed. Reactions that absorb energy from the surroundings are termed endothermic, while those that release energy are called exothermic. Another vital principle is the second law of thermodynamics, which introduces the concept of entropy—an important component of our current discussion.

Entropy is a measure of the disorder or randomness in a system. In chemical processes, we consider the entropy of the reactants and products to understand the direction in which a reaction is naturally inclined to proceed. The second law states that the total entropy of an isolated system can never decrease over time. This means that the entropy of the universe, which is the ultimate isolated system, tends to increase.
Entropy and Phase Changes
Entropy changes significantly during phase transitions. For instance, when a solid melts into a liquid or a liquid vaporizes into a gas, entropy increases because the molecules in the more disordered states (liquid and gas) have greater freedom of motion than in a solid.

An important rule of thumb is that, in general, the entropy of a substance increases with temperature or when a phase change occurs from solid to liquid to gas. However, it's not just about state changes; entropy can also change with chemical reactions or when a mix of substances changes its composition, as seen in the provided exercise.

In the context of chemical reactions, phase changes can have a significant influence on the system’s entropy. For instance, the production of gaseous products from solid reactants tends to lead to an entropy increase, as gas particles are more disordered than those in the solid phase. This is noticeable in the given exercise for reaction (b), where the transformation of calcium carbonate, a solid, to calcium oxide, another solid, and carbon dioxide, a gas, results in greater disorder and thus a positive entropy change.
Predicting Entropy Changes
Predicting the sign of entropy changes in chemical reactions involves assessing the disorder before and after the reaction. A reaction that results in more gas particles or a greater number of moles usually indicates an increase in entropy. In contrast, a reaction that yields fewer gas molecules or less volume will typically signify a decrease in entropy.

When we analyze the exercises provided, we can apply these concepts to predict the signs of entropy changes. For reaction (c), the entropy decreases because three moles of acetylene gas react to form only one mole of benzene gas, leading to a decrease in the system’s disorder. This illustrates well how a lessening in particle count, especially in gases, can signify a reduction in entropy.

However, it’s crucial to remember that the nature of reactants and products also matters. In reaction (d), we convert three gaseous molecules and one solid into two solids and three gases. Despite the moles of gas remaining constant, there is an overall increase in the number of particles, signifying increased disorder. Therefore, we would predict that the entropy change for reaction (d) is positive. Thus, practical experience and an understanding of molecular behavior in different states are essential in predicting entropy changes in chemical reactions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Consider the vaporization of liquid water to steam at a pressure of 1 atm. (a) Is this process endothermic or exothermic? (b) In what temperature range is it a spontaneous process? (c) In what temperature range is it a nonspontaneous process? (d) At what temperature are the two phases in equilibrium?

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?

A system goes from state 1 to state 2 and back to state 1 . (a) What is the relationship between the value of \(\Delta E\) for going from state 1 to state 2 to that for going from state 2 back to state \(1 ?\) (b) Without further information, can you conclude anything about the amount of heat transferred to the system as it goes from state 1 to state 2 as compared to that upon going from state 2 back to state \(1 ?(\mathrm{c})\) Suppose the changes in state are reversible processes. Can you conclude anything about the work done by the system upon going from state 1 to state 2 as compared to that upon going from state 2 back to state \(1 ?\)

About \(86 \%\) of the world's electrical energy is produced by using steam turbines, a form of heat engine. In his analysis of an ideal heat engine, Sadi Carnot concluded that the maximum possible efficiency is defined by the total work that could be done by the engine, divided by the quantity of heat available to do the work (for example, from hot steam produced by combustion of a fuel such as coal or methane). This efficiency is given by the ratio \(\left(T_{\text {high }}-T_{\text {low }}\right) / T_{\text {high }}\), where \(T_{\text {high }}\) is the temperature of the heat going into the engine and \(T_{\text {low }}\) is that of the heat leaving the engine. (a) What is the maximum possible efficiency of a heat engine operating between an input temperature of \(700 \mathrm{~K}\) and an exit temperature of \(288 \mathrm{~K} ?\) (b) Why is it important that electrical power plants be located near bodies of relatively cool water? (c) Under what conditions could a heat engine operate at or near \(100 \%\) efficiency? (d) It is often said that if the energy of combustion of a fuel such as methane were captured in an electrical fuel cell instead of by burning the fuel in a heat engine, a greater fraction of the energy could be put to useful work. Make a qualitative drawing like that in Figure 5.10 that illustrates the fact that in principle the fuel cell route will produce more useful work than the heat engine route from combustion of methane.

(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 \(19.12 ?\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.