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Indicate whether each statement is true or false. (a) Unlike enthalpy, where we can only ever know changes in \(H,\) we can know absolute values of \(S .(\mathbf{b})\) If you heat a gas such as \(\mathrm{CO}_{2}\), you will increase its degrees of translational, rotational and vibrational motions. (c) \(\mathrm{CO}_{2}(g)\) and \(\mathrm{Ar}(g)\) have nearly the same molar mass. At a given temperature, they will have the same number of microstates.

Short Answer

Expert verified
(a) True - We can know the absolute values of entropy due to the reference point provided by the third law of thermodynamics. (b) True - Heating a gas like CO2 increases its degrees of translational, rotational, and vibrational motions. (c) False - CO2(g) and Ar(g) have different numbers of microstates at a given temperature due to their differing molecular complexities.

Step by step solution

01

Statement (a) Evaluation

In this statement, the claim is that unlike enthalpy, we can know the absolute values of entropy. Enthalpy (H) is a state function, and we can only measure the change in enthalpy between two states (∆H). On the other hand, entropy (S) is also a state function, but we can determine its absolute value because we have a reference point, the entropy of a perfect crystal at 0 K, which is defined as zero according to the third law of thermodynamics. Therefore, this statement is \(True\).
02

Statement (b) Evaluation

This statement asserts that heating a gas like CO2 will increase its degrees of translational, rotational, and vibrational motions. As a gas molecule is heated, its energy increases leading to an increase in its different types of motion, which include translational, rotational, and vibrational motions. Therefore, this statement is \(True\).
03

Statement (c) Evaluation

The claim in this statement is that CO2(g) and Ar(g) have nearly the same molar mass, and at a given temperature, they will have the same number of microstates. While it may be true that CO2(g) and Ar(g) have nearly the same molar mass, the number of microstates depends not only on the molar mass but also on the molecule's complexity. Since CO2 is a more complex molecule compared to the noble gas Ar due to its additional vibrational and rotational modes, the number of microstates at a given temperature will be different between them. Therefore, this statement is \(False\).

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Key Concepts

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

Entropy
Entropy is a fascinating concept in thermodynamics that often puzzles students. It serves as a measure of disorder or randomness in a system. Imagine a room filled with toys haphazardly—a high-entropy situation. Now, think of the same room with toys neatly arranged; that's low entropy.
In thermodynamics, entropy is not just about disorder but about energy dispersal. When energy spreads out over more positions and states, entropy increases. Notably, entropy is a state function. This means that it depends only on the state of the system, not on how that state was achieved.
  • The absolute value of entropy can be determined thanks in part to the Third Law of Thermodynamics. This principle states that the entropy of a perfect crystal at absolute zero (0 K) is exactly zero. This gives us a baseline to calculate the absolute entropy of substances at different temperatures.
  • Therefore, unlike enthalpy where we only measure changes, entropy allows us an absolute measurement because we have a reference point at 0 Kelvin.
State Functions
In the world of thermodynamics, state functions are crucial for understanding how systems behave. A state function is a property whose value does not depend on the path taken to reach that specific value. This includes quantities like enthalpy, entropy, and internal energy.
Such functions provide valuable insights because they simplify calculations. For example, if you're trying to determine the change in temperature, you only need to consider its initial and final states, not every intermediate step.
This is particularly handy when analyzing processes like chemical reactions. You can focus on initial reactants and final products, skipping the messy path. Entropy, as we discussed, is a state function—its value relates to the state of a system and remains unaffected by path.
State functions play a significant role in evaluating the efficiency and feasibility of processes. They allow us to focus on essential aspects of energy transformations without getting bogged down in complexities of how these transformations occur.
Molecular Motion
Molecules are always in motion, and this motion increases with energy input. Such motions can be categorized into different types: translational, rotational, and vibrational.
  • Translational Motion: This type of motion translates a molecule from one place to another. Think of a gas particle moving across a room. When a gas is heated, its molecules move faster, increasing translational motion.
  • Rotational Motion: As its name suggests, this refers to molecules rotating around an axis. Gases, especially, enjoy more freedom to rotate than liquids or solids.
  • Vibrational Motion: In this type of motion, the atoms within a molecule oscillate about their positions. In gases, vibrational motion becomes more significant as the temperature rises.
When you heat a gas like CO2, you add energy, increasing its molecular motion in all these forms. Complex molecules like CO2 have more modes of motion compared to simple ones like Ar, influencing their entropy and the number of accessible microstates.

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Most popular questions from this chapter

The conversion of natural gas, which is mostly methane, into products that contain two or more carbon atoms, such as ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\), is a very important industrial chemical process. In principle, methane can be converted into ethane and hydrogen: $$ 2 \mathrm{CH}_{4}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)+\mathrm{H}_{2}(g) $$ In practice, this reaction is carried out in the presence of oxygen: $$ 2 \mathrm{CH}_{4}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$ (a) Using the data in Appendix \(C\), calculate \(K\) for these reactions at \(25^{\circ} \mathrm{C}\) and \(500^{\circ} \mathrm{C}\). (b) Is the difference in \(\Delta G^{\circ}\) for the two reactions due primarily to the enthalpy term \((\Delta H)\) or the entropy term \((-T \Delta S)\) ? (c) Explain how the preceding reactions are an example of driving a nonspontaneous reaction, as discussed in the "Chemistry and Life" box in Section 19.7. (d) The reaction of \(\mathrm{CH}_{4}\) and \(\mathrm{O}_{2}\) to form \(\mathrm{C}_{2} \mathrm{H}_{6}\) and \(\mathrm{H}_{2} \mathrm{O}\) must be carried out carefully to avoid a competing reaction. What is the most likely competing reaction?

Does the entropy of the system increase, decrease, or stay the same when (a) the temperature of the system increases, (b) the volume of a gas increases, (c) equal volumes of ethanol and water are mixed to form a solution?

Carbon disulfide \(\left(C S_{2}\right)\) is a toxic, highly flammable substance. The following thermodynamic data are available for \(\mathrm{CS}_{2}(I)\) and \(\mathrm{CS}_{2}(g)\) at \(298 \mathrm{~K}\) \begin{tabular}{lcc} \hline & \(\Delta H_{i}(\mathrm{k} / \mathrm{mol})\) & \(\Delta G_{i}^{\prime}(\mathrm{kJ} / \mathrm{mol})\) \\ \hline\(C S_{2}(l)\) & 89.7 & 65.3 \\ \(C S_{2}(g)\) & 117.4 & 67.2 \\ \hline \end{tabular} (a) Draw the Lewis structure of the molecule. What do you predict for the bond order of the \(\mathrm{C}-\mathrm{S}\) bonds? \((\mathbf{b})\) Use the VSEPR method to predict the structure of the \(\mathrm{CS}_{2}\) molecule. (c) Liquid \(\mathrm{CS}_{2}\) burns in \(\mathrm{O}_{2}\) with a blue flame, forming \(\mathrm{CO}_{2}(g)\) and \(\mathrm{SO}_{2}(g)\). Write a balanced equation for this reaction. (d) Using the data in the preceding table and in Appendix \(C,\) calculate \(\Delta H^{\circ}\) and \(\Delta G^{\circ}\) for the reaction in part \((c) .\) Is the reaction exothermic? Is it spontaneous at \(298 \mathrm{~K} ?\) (e) Use the data in the table to calculate \(\Delta S^{\circ}\) at \(298 \mathrm{~K}\) for the vaporization of \(\mathrm{CS}_{2}(I) .\) Is the sign of \(\Delta S^{\circ}\) as you would expect for a vaporization? (f) Using data in the table and your answer to part (e), estimate the boiling point of \(\mathrm{CS}_{2}(l)\). Do you predict that the substance will be a liquid or a gas at \(298 \mathrm{~K}\) and \(101.3 \mathrm{kPa}\) ?

A certain reaction has \(\Delta H^{\circ}=+20.0 \mathrm{~kJ}\) and \(\Delta S^{\circ}=\) \(+100.0 \mathrm{~J} / \mathrm{K} .\) (a) Does the reaction lead to an increase or decrease in the randomness or disorder of the system? (b) Does the reaction lead to an increase or decrease in the randomness or disorder of the surroundings? (c) Calculate \(\Delta G^{\circ}\) for the reaction at \(298 \mathrm{~K} .(\mathbf{d})\) Is the reaction spontaneous at \(298 \mathrm{~K}\) under standard conditions?

Predict which member of each of the following pairs has the greater standard entropy at \(25^{\circ} \mathrm{C}:\) (a) \(\mathrm{C}_{6} \mathrm{H}_{6}(l)\) or \(\mathrm{C}_{6} \mathrm{H}_{6}(g)\), (b) \(\mathrm{CO}(g)\) or \(\mathrm{CO}_{2}(g)\) (c) \(1 \mathrm{~mol} \mathrm{~N}_{2} \mathrm{O}_{4}(g)\) or \(2 \mathrm{~mol} \mathrm{NO}_{2}(g)\) (d) \(\mathrm{HCl}(g)\) or \(\mathrm{HCl}(a q) .\) Use Appendix \(\mathrm{C}\) to find the standard entropy of each substance.

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