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Predict which member of each of the following pairs has the greater standard entropy at \(25^{\circ} \mathrm{C} :(\mathbf{a}) \operatorname{Sc}(s)\) or \(\operatorname{Sc}(g)\) (b) \(\mathrm{NH}_{3}(g)\) or \(\mathrm{NH}_{3}(a q),(\mathbf{c}) \mathrm{O}_{2}(g)\) or \(\mathrm{O}_{3}(g),(\mathbf{d}) \mathrm{C}(\mathrm{graphite})\) or \(\mathrm{C}(\) diamond). Use Appendix \(\mathrm{C}\) to find the standard entropy of each substance.

Short Answer

Expert verified
In conclusion, the members with greater standard entropies at \(25^{\circ}\mathrm{C}\) for each pair are: a) Sc(g) with 182.9 J/mol·K b) NH3(g) with 192.5 J/mol·K c) O3(g) with 238.9 J/mol·K d) C(graphite) with 5.74 J/mol·K

Step by step solution

01

Pair A: Sc(s) or Sc(g)

To compare the standard entropies of Scandium (Sc) in solid and gaseous states, look up the values in Appendix C. Typically, gases have higher standard entropies than solids due to their higher degrees of freedom, so we expect Sc(g) to have a greater standard entropy than Sc(s). Let's verify this using the given data. From Appendix C, Standard Entropy of Sc(s) = 14.88 J/mol·K Standard Entropy of Sc(g) = 182.9 J/mol·K Comparing these values, we can see that Sc(g) has a greater standard entropy than Sc(s).
02

Pair B: NH3(g) or NH3(aq)

Next, we will compare the standard entropies of ammonia (NH3) in gaseous state and aqueous solution. Gases usually have higher standard entropies than liquids or solutes in solutions due to the greater freedom of movement, so NH3(g) should have a greater standard entropy than NH3(aq). Let's check this using values from Appendix C. From Appendix C, Standard Entropy of NH3(g) = 192.5 J/mol·K Standard Entropy of NH3(aq) = 111.3 J/mol·K Comparing these values, we can see that NH3(g) has a greater standard entropy than NH3(aq).
03

Pair C: O2(g) or O3(g)

Now we need to compare the standard entropies of oxygen gas (O2) and ozone gas (O3). As both are gases, their standard entropies may not differ as much, but larger molecules tend to have greater standard entropies due to their enhanced complexity. Let's look up the values in Appendix C to compare them. From Appendix C, Standard Entropy of O2(g) = 205.1 J/mol·K Standard Entropy of O3(g) = 238.9 J/mol·K Comparing these values, we can see that O3(g) has a greater standard entropy than O2(g).
04

Pair D: C(graphite) or C(diamond)

Finally, we will compare the standard entropies of carbon in graphite and diamond forms. As both are solid, their standard entropies might be similar. However, graphite has a more disordered structure than diamond, which might lead to a higher standard entropy. Let's verify this by checking the values in Appendix C. From Appendix C, Standard Entropy of C(graphite) = 5.74 J/mol·K Standard Entropy of C(diamond) = 2.52 J/mol·K Comparing these values, we can see that C(graphite) has a greater standard entropy than C(diamond). In conclusion, the members with greater standard entropies for each pair are: a) Sc(g) b) NH3(g) c) O3(g) d) C(graphite)

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Key Concepts

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

Entropy of Solids and Gases
Entropy, a key thermodynamic property, can be thought of as a measure of disorder or randomness in a system. For students tackling textbook exercises, it's helpful to know that the entropy of a substance depends strongly on its state of matter. Solids, like scandium (Sc) in solid form, tend to have lower entropy than gases because their particles are closely packed and have limited freedom of movement. This is why Sc(s) has a significantly lower entropy value of 14.88 J/mol·K compared to Sc(g) at 182.9 J/mol·K.

When comparing gases and solids, remember that gas particles move freely and rapidly, filling the container's volume, which results in a higher degree of randomness and, consequently, higher entropy. Understanding this difference is crucial for predicting which substance in various states has greater entropy.
Standard Entropy Comparison
Standard entropy values are tabulated under standard conditions, which is 25°C and 1 atmosphere of pressure for most substances. Comparing these values, as seen in the exercise, is straightforward when you have a solid understanding of the concept of entropy. The standard entropy of NH3(g) at 192.5 J/mol·K is higher than that of NH3(aq) at 111.3 J/mol·K, reflecting the difference between the gaseous state and the higher order in a solution. Similarly, the entropy of larger molecules like O3(g) at 238.9 J/mol·K is greater than that of simpler molecules like O2(g) at 205.1 J/mol·K because of the increased complexity and degrees of freedom. Applying this knowledge helps to easily determine the substance with the higher standard entropy in a pair.
Entropy in Different States of Matter
As students advance in understanding entropy, they learn that it is not just the state of matter—solid, liquid, or gas—that affects entropy, but also the molecular complexity and structure. For example, all else being equal, liquids generally have lower entropy than gases. However, within the solid state, substances like graphite and diamond demonstrate that molecular arrangement impacts entropy. Graphite's looser structure, with layers that can slide past one another, endows it with a higher entropy (5.74 J/mol·K) than diamond's rigid, tightly-bonded lattice (2.52 J/mol·K). Through exercises like these, students can reinforce their understanding that entropy is context-dependent and varies with molecular detail and state of matter.
Thermodynamic Properties of Substances
The thermodynamic properties of substances, which include not only entropy but also enthalpy, free energy, and specific heat, are fundamental to understand chemical behavior. Thermodynamics tells us how energy is transferred in reactions and how it affects substance properties. Entropy is just one part of this bigger picture, reflecting the degree of disorder. By looking at standard entropy values, students can infer a lot about substance behavior. For example, the greater entropy of Sc(g) over Sc(s) suggests it is more energetically favorable for scandium to be in a gaseous state at standard conditions. Recognizing these properties and their interrelations allows students to predict reaction spontaneity and equilibrium positions, key aspects of chemical thermodynamics.

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Most popular questions from this chapter

Consider a system consisting of an ice cube. (a) Under what conditions can the ice cube melt reversibly? If the ice cube melts reversibly, is \(\Delta H\) zero for the process?

Calculate \(\Delta S^{\circ}\) values for the following reactions by using tabulated \(S^{\circ}\) values from Appendix \(\mathrm{C} .\) In each case, explain the sign of \(\Delta S^{\circ} .\) $$ \begin{array}{l}{\text { (a) } \mathrm{HNO}_{3}(g)+\mathrm{NH}_{3}(g) \longrightarrow \mathrm{NH}_{4} \mathrm{NO}_{3}(s)} \\ {\text { (b) } 2 \mathrm{Fe}_{2} \mathrm{O}_{3}(s) \longrightarrow 4 \mathrm{Fe}(s)+3 \mathrm{O}_{2}(g)} \\ {\text { (c) } \mathrm{CaCO}_{3}(s, \text { calcite })+2 \mathrm{HCl}(g) \rightarrow} \\\ {\mathrm{CaCl}_{2}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)}\\\ {\text { (d) } 3 \mathrm{C}_{2} \mathrm{H}_{6}(g) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{6}(l)+6 \mathrm{H}_{2}(g)}\end{array} $$

Which of the following processes are spontaneous and which are nonspontaneous: (a) the ripening of a banana, (b) dissolution of sugar in a cup of hot coffee, (c) the reaction of nitrogen atoms to form \(\mathrm{N}_{2}\) molecules at \(25^{\circ} \mathrm{C}\) and \(1 \mathrm{atm},(\mathbf{d})\) lightning, (e) formation of \(\mathrm{CH}_{4}\) and \(\mathrm{O}_{2}\) molecules from \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) at room temperature and 1 atm of pressure?

A system goes from state 1 to state 2 and back to state \(1 .\) (a) Is \(\Delta E\) the same in magnitude for both the forward and reverse processes? (b) Without further information, can you conclude that the amount of heat transferred to the system as it goes from state 1 to state 2 is the same or different as compared to that upon going from state 2 back to state 1\(?(\mathbf{c})\) Suppose the changes in state are reversible processes. Is the work done by the system upon going from state 1 to state 2 the same or different as compared to that upon going from state 2 back to state 1\(?\)

The reaction $$ \mathrm{SO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{S}(g) \rightleftharpoons 3 \mathrm{S}(s)+2 \mathrm{H}_{2} \mathrm{O}(g) $$ is the basis of a suggested method for removal of SO \(_{2}\) from power-plant stack gases. The standard free energy of each substance is given in Appendix C. (a) What is the equilibrium constant for the reaction at 298 \(\mathrm{K} ?\) (b) In principle, is this reaction a feasible method of removing \(S O_{2} ?\) (c) If \(P_{S O_{2}}=P_{H_{2} S}\) and the vapor pressure of water is 25 torr, calculate the equilibrium \(\mathrm{SO}_{2}\) pressure in the system at 298 K. (d) Would you expect the process to be more or less effective at higher temperatures?

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