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Chapter 19: Problem 86

The crystalline hydrate \(\mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2} \cdot 4 \mathrm{H}_{2} \mathrm{O}(s)\) loses water when placed in a large, closed, dry vessel at room temperature: $$ \mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2} \cdot 4 \mathrm{H}_{2} \mathrm{O}(s) \longrightarrow \mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2}(s)+4 \mathrm{H}_{2} \mathrm{O}(g) $$ This process is spontaneous and \(\Delta H^{\circ}\) is positive at room temperature. (a) What is the sign of \(\Delta S^{\circ}\) at room temperature? (b) If the hydrated compound is placed in a large, closed vessel that already contains a large amount of water vapor, does \(\Delta S^{\circ}\) change for this reaction at room temperature?

Short Answer

Expert verified
(a) \(\Delta S^{\circ}\) is positive at room temperature. (b) No, placing the hydrated compound in a large, closed vessel already containing a large amount of water vapor does not change the intrinsic entropy change (\(\Delta S^{\circ}\)) of the reaction at room temperature.

Step by step solution

01

Calculate change in entropy \(\Delta S^{\circ}\) and its sign at room temperature

The relationship between Gibbs free energy (\(\Delta G^{\circ}\)), enthalpy (\(\Delta H^{\circ}\)), and entropy (\(\Delta S^{\circ}\)) is given by the equation: \[\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}\] Since the process is spontaneous, the change in Gibbs free energy (\(\Delta G^{\circ}\)) must be negative. We are given that the change in enthalpy (\(\Delta H^{\circ}\)) is positive for this process. The only way to satisfy a negative value for \(\Delta G^{\circ}\) is if the term \(T\Delta S^{\circ}\) is greater in magnitude than \(\Delta H^{\circ}\) and therefore has a positive value: \[\Delta H^{\circ} - T\Delta S^{\circ} < 0 \Rightarrow T\Delta S^{\circ} > \Delta H^{\circ}\] Since room temperature is a positive value, the change in entropy (\(\Delta S^{\circ}\)) must also be positive to satisfy the inequality. Thus, we can conclude that: (a) \(\Delta S^{\circ}\) is positive at room temperature.
02

Effect of additional water vapor on \(\Delta S^{\circ}\) at room temperature

Placing the hydrated compound in a large, closed vessel already containing a large amount of water vapor will not change the intrinsic change in entropy (\(\Delta S^{\circ}\)) as \(\Delta S^{\circ}\) depends on the reaction itself and not on the initial conditions of the system. However, it can potentially affect partial pressures of the water below and above the hydrate, which in turn can affect the spontaneity of the process. Nevertheless, the intrinsic change in entropy at room temperature for the reaction will remain the same. Therefore, we can conclude that: (b) No, placing the hydrated compound in a large, closed vessel already containing a large amount of water vapor does not change the intrinsic entropy change (\(\Delta S^{\circ}\)) of the reaction at room temperature.

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Key Concepts

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

Spontaneity
The spontaneity of a reaction refers to whether a process will occur naturally without any external influence. A spontaneous process is one that can proceed on its own. An important component to determine spontaneity is the Gibbs free energy change, represented as \( \Delta G^{\circ} \).
- If \( \Delta G^{\circ} \) is negative, the reaction is spontaneous.- If \( \Delta G^{\circ} \) is positive, the reaction is non-spontaneous.In the example of the crystalline hydrate losing water, the process is noted as spontaneous, meaning it releases water vapor without requiring additional energy or changes. This natural shift, despite having a positive enthalpy change, is still favorable because the entropy change \( \Delta S^{\circ} \) is significant enough to drive the reaction.forward.
Enthalpy
Enthalpy, represented as \( \Delta H^{\circ} \), is a measure of the total heat content of a system. Changes in enthalpy relate to the heat absorbed or released due to a reaction at constant pressure.
- A positive \( \Delta H^{\circ} \) means the reaction absorbs heat, also known as an endothermic reaction.- A negative \( \Delta H^{\circ} \) indicates the release of heat, known as an exothermic reaction.In the crystalline hydrate example, the enthalpy change is positive, signifying that the process of releasing water vapor absorbs heat. Despite this, because the process is spontaneous, another factor鈥攍ike a significant increase in entropy鈥攃ompensates for the positive \( \Delta H^{\circ} \). This dissociation highlights that enthalpy alone doesn't determine spontaneity, but must be considered along with entropy and temperature.
Gibbs Free Energy
Gibbs free energy is a crucial thermodynamic quantity, denoted as \( \Delta G^{\circ} \), that dictates the spontaneity of a reaction at constant temperature and pressure. It combines enthalpy, temperature, and entropy through the formula:\[ \Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ} \]
The equation shows how temperature and entropy impact the Gibbs free energy and ultimately the spontaneity:- Lower \( \Delta G^{\circ} \) (i.e., negative) suggests that a reaction will occur spontaneously.In the example of the crystalline hydrate, even with a positive \( \Delta H^{\circ} \), the process remains spontaneous. This indicates that the \( T\Delta S^{\circ} \) term is sufficiently large in positive value, making \( \Delta G^{\circ} \) negative. This interplay highlights how reactions that absorb heat can still proceed naturally if they result in increases in disorder or randomness (entropy). Thus, Gibbs free energy effectively predicts the feasibility of reactions in varying conditions.

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Most popular questions from this chapter

(a) For each of the following reactions, predict the sign of \(\Delta H^{*}\) and \(\Delta S^{\circ}\) without doing any calculations. (b) Based on your general chemical knowledge, predict which of these reactions will have \(K>1\) at \(25^{\circ} \mathrm{C} .(\mathbf{c})\) In each case, indicate whether \(K\) should increase or decrease with increasing temperature. (i) \(2 \mathrm{Fe}(s)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{FeO}(s)\) (ii) \(\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{Cl}(g)\) (iii) \(\mathrm{NH}_{4} \mathrm{Cl}(s) \rightleftharpoons \mathrm{NH}_{3}(g)+\mathrm{HCl}(g)\) (iv) \(\mathrm{CO}_{2}(\mathrm{~g})+\mathrm{CaO}(s) \rightleftharpoons \mathrm{CaCO}_{3}(s)\)

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 \(\mathrm{SO}_{2}\) 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 \(298 \mathrm{~K} ?(\mathbf{b})\) In principle, is this reaction a feasible method of removing \(\mathrm{SO}_{2}\) ? (c) If \(P_{5 \mathrm{O}_{2}}=P_{\mathrm{H}_{2}}\) s and the vapor pressure of water is \(3.33 \mathrm{kPa}\), calculate the equilibrium \(\mathrm{SO}_{2}\) pressure in the system at \(298 \mathrm{~K}\). (d) Would you expect the process to be more or less effective at higher temperatures?

Predict which member of each of the following pairs has the greater standard entropy at \(25^{\circ} \mathrm{C}:\) (a) \(\mathrm{HNO}_{3}(g)\) or \(\mathrm{HNO}_{3}(a q)\) (b) \(\mathrm{PCl}_{3}(l)\) or \(\mathrm{PCl}_{3}(g)\), (c) \(\mathrm{Fe}_{2} \mathrm{O}_{3}(s)\) or \(\mathrm{Fe}_{3} \mathrm{O}_{4}(s),(\mathbf{d}) \mathrm{Li}(s)\) or \(\mathrm{Li}(g)\). Use Appendix \(\mathrm{C}\) to find the stan- dard entropy of each substance.

Consider the reaction \(\mathrm{CH}_{4}(\mathrm{~g})+4 \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{CCl}_{4}(g)+\) \(4 \mathrm{HCl}(g) .\). (a) Using data from Appendix C, calculate \(\Delta G^{\circ}\) at \(298 \mathrm{~K} .(\mathbf{b})\) Calculate \(\Delta G\) at \(298 \mathrm{~K}\) if the reaction mixture consists of \(50.7 \mathrm{kPa}\) of \(\mathrm{CH}_{4}(g), 25.3 \mathrm{kPa}\) of \(\mathrm{Cl}_{2}(g), 10.13 \mathrm{kPa}\) of \(\mathrm{CCl}_{4}(\mathrm{~g})\) and \(15.2 \mathrm{kPa}\) of \(\mathrm{HCl}(\mathrm{g})\)

A standard air conditioner involves a \(r\) frigerant that is typically now a fluorinated hydrocarbon, such as \(\mathrm{CH}_{2} \mathrm{~F}_{2}\). An air- conditioner refrigerant has the property that it readily vaporizes at atmospheric pressure and is easily compressed to its liquid phase under increased pressure. The operation of an air conditioner can be thought of as a closed system made up of the refrigerant going through the two stages shown here (the air circulation is not shown in this diagram). During expansion, the liquid refrigerant is released into an expansion chamber at low pressure, where it vaporizes. The vapor then undergoes compression at high pressure back to its liquid phase in a compression chamber. (a) What is the sign of \(q\) for the expansion? (b) What is the sign of \(q\) for the compression? (c) In a central air-conditioning system, one chamber is inside the home and the other is outside. Which chamber is where, and why? (d) Imagine that a sample of liquid refrigerant undergoes expansion followed by compression, so that it is back to its original state. Would you expect that to be a reversible process? (e) Suppose that a house and its exterior are both initially at \(31^{\circ} \mathrm{C}\). Some time after the air conditioner is turned on, the house is cooled to \(24^{\circ} \mathrm{C}\). Is this process spontaneous of nonspontaneous?
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