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For each of the following pairs of substances, which substance has the greater value of \(S^{\circ}\) ? a. \(\mathrm{C}_{\text {graphite }}(s)\) or \(\mathrm{C}_{\text {diamond }}(s)\) b. \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)\) or \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(g)\) c. \(\mathrm{CO}_{2}(s)\) or \(\mathrm{CO}_{2}(g)\)

Short Answer

Expert verified
a. C(graphite)(s) has the greater value of S°. b. C2H5OH(g) has the greater value of S°. c. CO2(g) has the greater value of S°.

Step by step solution

01

Case a: C(graphite)(s) or C(diamond)(s)

In this case, we are comparing the entropy of graphite and diamond, which are both forms of solid carbon. Between these two, graphite has a more disordered structure as its carbon atoms are arranged in sheets with weak inter-sheet bonding. In contrast, diamond has a very ordered and strong three-dimensional structure. Thus, graphite should have a higher standard molar entropy than diamond.
02

Case b: C2H5OH(l) or C2H5OH(g)

We are comparing the entropy of liquid ethanol and gaseous ethanol. In general, the entropy of a substance increases as it changes from solid to liquid to gas. This is because the particles are more loosely bonded and free to move in the gas phase compared to the liquid phase. Therefore, the standard molar entropy of gaseous ethanol (C2H5OH(g)) is greater than that of liquid ethanol (C2H5OH(l)).
03

Case c: CO2(s) or CO2(g)

Similarly, we are comparing the entropy of solid carbon dioxide (also known as dry ice) and gaseous carbon dioxide. As mentioned earlier, the entropy of a substance increases in the order of solid, liquid, and gas due to the increasing level of disorder. In this case, the gaseous phase (CO2(g)) has a higher standard molar entropy than the solid phase (CO2(s)).

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Key Concepts

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

Standard Molar Entropy (S^o)
Standard molar entropy, denoted as \(S^{\circ}\), is a way to quantify the degree of disorder or randomness associated with one mole of a substance at a standard reference state. The units for standard molar entropy are typically joules per mole kelvin (J/(mol·K)). Greater entropy values suggest a higher level of disorder and a greater dispersal of energy within the system.

In understanding \(S^{\circ}\), it's important to note that the entropy of a substance is intrinsically linked to the freedom of its particles. As a general rule, gases have higher entropy than liquids, which in turn have higher entropy than solids. This is because particles in a gas have the most freedom to move and occupy space compared to particles in liquids and solids. Consequently, predicting which substance has greater \(S^{\circ}\) often involves considering its physical state under standard conditions as well as the complexity of its molecular structure.
Entropy of Carbon Allotropes
Carbon allotropes such as graphite and diamond exhibit different structures and, thus, different standard molar entropies. Graphite consists of layers of carbon atoms arranged in a hexagonal lattice, where each layer is weakly bonded to the next. This structure allows for more randomness in the arrangement of atoms and easier movement within the layers.

On the other hand, diamond features a rigid, tetrahedral lattice where each carbon atom is strongly bonded to four others, forming a very stable and highly ordered structure. The high order and strong bonding in diamond minimize the disorder and possible microstates, leading to a lower entropy in comparison to graphite. In summary, between graphite and diamond, graphite has the higher standard molar entropy due to its more disordered structure.
Phases of Matter
The phases or states of matter—solid, liquid, and gas—are differentiated by the degree of order and strength of interaction among their constituent particles. In solids, particles are closely packed in a fixed volume and shape, exhibiting vibrations but minimal movement. Liquids have more freedom, allowing particles to slide past one another while generally maintaining a fixed volume. Gases possess the most freedom, with particles that can move independently and completely fill their container.

Increasing Entropy Across Phases

Order decreases significantly from solids to liquids to gases, which also correlates with increasing entropy. The sequence of phases—solid to liquid to gas—is thus a sequence of increasing entropy. When a substance transitions from a solid to a liquid to a gas, the particles become less ordered and more widely distributed, leading to higher randomness, more possible microstates, and ultimately, higher entropy.
Entropy Changes in Phase Transitions
Phase transitions, such as melting or vaporizing, involve significant changes in the properties of a substance, including its entropy. Consider the transition from a solid to a liquid (melting) or from a liquid to a gas (boiling).

Melting and Boiling

During melting, the solid structure breaks down, allowing particles to move past one another more freely in the liquid state. The increased motion and disorder cause a rise in entropy. Boiling presents an even more drastic change—particles in a liquid are confined to a certain volume, but upon vaporization, they disperse throughout the available space, greatly increasing the substance's entropy.

Entropy changes are also reversible. For example, when a gas condenses into a liquid, or a liquid freezes into a solid, entropy decreases because the particles are becoming more ordered and less dispersed. Thus, a substance's phase at a particular temperature and pressure can significantly influence its standard molar entropy.

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Most popular questions from this chapter

The equilibrium constant for a certain reaction increases by a factor of \(6.67\) when the temperature is increased from \(300.0 \mathrm{~K}\) to \(350.0 \mathrm{~K} .\) Calculate the standard change in enthalpy \(\left(\Delta H^{\circ}\right)\) for this reaction (assuming \(\Delta H^{\circ}\) is temperature-independent).

It is quite common for a solid to change from one structure to another at a temperature below its melting point. For example, sulfur undergoes a phase change from the rhombic crystal structure to the monoclinic crystal form at temperatures above \(95^{\circ} \mathrm{C}\). a. Predict the signs of \(\Delta H\) and \(\Delta S\) for the process \(\mathrm{S}_{\text {rhombic }}(s) \longrightarrow \mathrm{S}_{\text {monoclinic }}(s)\) b. Which form of sulfur has the more ordered crystalline structure (has the smaller positional probability)?

Predict the sign of \(\Delta S\) for each of the following and explain. a. the evaporation of alcohol b. the freezing of water c. compressing an ideal gas at constant temperature d. dissolving \(\mathrm{NaCl}\) in water

Impure nickel, refined by smelting sulfide ores in a blast furnace, can be converted into metal from \(99.90 \%\) to \(99.99 \%\) purity by the Mond process. The primary reaction involved in the Mond process is $$ \mathrm{Ni}(s)+4 \mathrm{CO}(g) \rightleftharpoons \mathrm{Ni}(\mathrm{CO})_{4}(g) $$ a. Without referring to Appendix 4 , predict the sign of \(\Delta S^{\circ}\) for the above reaction. Explain. b. The spontaneity of the above reaction is temperaturedependent. Predict the sign of \(\Delta S_{\text {surr }}\) for this reaction. Explain. c. For \(\mathrm{Ni}(\mathrm{CO})_{4}(g), \Delta H_{\mathrm{f}}^{\circ}=-607 \mathrm{~kJ} / \mathrm{mol}\) and \(S^{\circ}=417 \mathrm{~J} / \mathrm{K}\). mol at \(298 \mathrm{~K}\). Using these values and data in Appendix 4 , calculate \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for the above reaction. d. Calculate the temperature at which \(\Delta G^{\circ}=0(K=1)\) for the above reaction, assuming that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature. e. The first step of the Mond process involves equilibrating impure nickel with \(\mathrm{CO}(g)\) and \(\mathrm{Ni}(\mathrm{CO})_{4}(g)\) at about \(50^{\circ} \mathrm{C}\). The purpose of this step is to convert as much nickel as possible into the gas phase. Calculate the equilibrium constant for the above reaction at \(50 .{ }^{\circ} \mathrm{C}\). f. In the second step of the Mond process, the gaseous \(\mathrm{Ni}(\mathrm{CO})_{4}\) is isolated and heated to \(227^{\circ} \mathrm{C}\). The purpose of this step is to deposit as much nickel as possible as pure solid (the reverse of the preceding reaction). Calculate the equilibrium constant for the preceding reaction at \(227^{\circ} \mathrm{C}\). g. Why is temperature increased for the second step of the Mond process? h. The Mond process relies on the volatility of \(\mathrm{Ni}(\mathrm{CO})_{4}\) for its success. Only pressures and temperatures at which \(\mathrm{Ni}(\mathrm{CO})_{4}\) is a gas are useful. A recently developed variation of the Mond process carries out the first step at higher pressures and a temperature of \(152^{\circ} \mathrm{C}\). Estimate the maximum pressure of \(\mathrm{Ni}(\mathrm{CO})_{4}(g)\) that can be attained before the gas will liquefy at \(152^{\circ} \mathrm{C}\). The boiling point for \(\mathrm{Ni}(\mathrm{CO})_{4}\) is \(42^{\circ} \mathrm{C}\) and the enthalpy of vaporization is \(29.0 \mathrm{~kJ} / \mathrm{mol}\). [Hint: The phase change reaction and the corresponding equilibrium expression are $$ \mathrm{Ni}(\mathrm{CO})_{4}(l) \rightleftharpoons \mathrm{Ni}(\mathrm{CO})_{4}(g) \quad K=P_{\mathrm{Ni}(\mathrm{CO})} $$ \(\mathrm{Ni}(\mathrm{CO})_{4}(g)\) will liquefy when the pressure of \(\mathrm{Ni}(\mathrm{CO})_{4}\) is greater than the \(K\) value.]

Consider the reaction: $$ \mathrm{H}_{2} \mathrm{~S}(g)+\mathrm{SO}_{2}(g) \longrightarrow 3 \mathrm{~S}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) $$ for which \(\Delta H\) is \(-233 \mathrm{~kJ}\) and \(\Delta S\) is \(-424 \mathrm{~J} / \mathrm{K}\). a. Calculate the free energy change for the reaction \((\Delta G)\) at \(393 \mathrm{~K}\). b. Assuming \(\Delta H\) and \(\Delta S\) do not depend on temperature, at what temperatures is this reaction spontaneous?

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